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Transverse axis parallel to the x-axis,center at (2, -2) and passing through (2+3...

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bluecute | Student, Undergraduate | (Level 1) eNoter

Posted May 11, 2012 at 4:26 AM via web

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Transverse axis parallel to the x-axis,center at (2, -2) and passing through (2+3 square root of 2, 0) and (2+3 square root of 10, 4). Find the equation of hyperbola.

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thilina-g | College Teacher | (Level 1) Educator

Posted May 11, 2012 at 7:16 AM (Answer #1)

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If the transverse axis is parallel to x-axis or horizontal, the form of the equation of hyperbola is,

`(x-h)^2/a^2 - (y-k)^2/b^2 = 1`

(The transverse axis of a hyperbola is the line which connects the two vertices through the center of the hyperbola)

(Therefore this is an east-west opening hyperbola)

Where, (h,k) is the center of the hyperbola, a is the semi major axis and b is the semi minor axis.

It is given that the center of the hyperbola is (2,-2). Therefore, h = 2 and k = -2. The equations becomes,

`(x-2)^2/a^2 - (y-(-2))^2/b^2 = 1`

`(x-2)^2/a^2 - (y+2)^2/b^2 = 1`

To find a and b, we can incorporate the other two data given.

It passes through `(2+3sqrt(2),0)` and `(2+3sqrt(10),4)` . Using above two coordinates, we can find two equations for `1/a^2` and `1/b^2` .

Using first coordinate, `(2+3sqrt(2),0)`,

`(2+3sqrt(2)-2)^2/a^2 - (0+2)^2/b^2 = 1`

`(3sqrt(2))^2/a^2 - (2)^2/b^2 = 1`

`18/a^2 - 4/b^2 = 1`

This is equation 1.

Using second coordinate, `(2+3sqrt(10),4)`

`(2+3sqrt(10)-2)^2/a^2 - (4+2)^2/b^2 = 1`

`(3sqrt(10))^2/a^2 - (6)^2/b^2 = 1`

`90/a^2 - 36/b^2 = 1`

This is equation 2,

Now we have two equations,

   `18/a^2 - 4/b^2 = 1`  and   `90/a^2 - 36/b^2 = 1`

By solving these two simultaneous equations, you would get,

`1/a^2 = 1/9` and `1/b^2 = 1/4`

Therefore the equation of the hyperbola is,

`(x-2)^2/9 - (y+2)^2/4 = 1`

Or

`(x-2)^2/3^2 - (y+2)^2/2^2 = 1`

 

The center is at (2,-2), semi major axis is 3 and semi minor axis is 2.

 

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